To answer this question, we must noted that:
The slopes of two perpendicular lines are negative reciprocals of each other:
[tex]\begin{gathered} m_1m_2=-1 \\ \text{Where } \\ m_1\text{ is the slope of one of the lines} \\ m_2\text{ is the slope of the line perpendicular to the other line} \end{gathered}[/tex]
First, we will find what the slope of the initial equation is:
[tex]\begin{gathered} 10x-8y=-80 \\ 10x+80=8y \\ \frac{10x}{8}+\frac{80}{8}=\frac{8y}{8} \\ \frac{5}{4}x+10=y \end{gathered}[/tex]
So from the above solution, we see that:
[tex]m_1=\frac{5}{4}[/tex]
We will solve for the slope of a line perpendicular to it thus:
[tex]\begin{gathered} m_1m_2=-1 \\ \frac{5}{4}m_2=-1 \end{gathered}[/tex]
Simplifying further:
[tex]\begin{gathered} m_2=-\frac{1}{\frac{5}{4}} \\ =-1\div\frac{5}{4} \\ =-1\times\frac{4}{5} \\ =\frac{-4}{5} \\ \text{The slope of a line perpendicular to the line with equation 10x-8y=-80 is:} \\ \frac{-4}{5} \end{gathered}[/tex]
